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I have $r^2=-4\sinθ$

and I'm asked to set $r=0$, then find θ. If I just set $r^2=0$ then I'll get $\sin(2θ)=0$. That doesn't seem right.

Then I'm asked to set $θ=0$ and then find $r$. If I use the $r^2=-4\sinθ$ and set $θ=0$ then I will get "DNE". Not sure what to do instead then...


Sorry everyone I wrote the problem here wrong. It was supposed to be r^2=-4sin2θ

That's where the 2θ came from

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Are you doing this all on a calculator? What do you know about the value of $\sin(0)$? What do you know about $\sin(2\pi)$? No calculator needed here. – amWhy May 21 '11 at 21:34
Where'd $\sin(2\theta)$ come from? – Hans Parshall May 21 '11 at 21:37
It is not clear what the question is. But looking at $r^2=-4\sin\theta$, the first thing I would notice is that since $r^2 \ge 0$, the curve makes sense only when $\sin\theta \le 0$, so, in the interval from $0$ to $2\pi$, only when $\pi \le \theta \le 2\pi$. – André Nicolas May 21 '11 at 21:44
You've got three questions going on the same problem! That's not a good use of the available resources. – Gerry Myerson May 22 '11 at 6:17
up vote 3 down vote accepted

For any real number, $r=0$ if and only if $r^2=0$, so "set[ting] $r=0$" is the same as setting $r^2$ to zero. Equivalently: if $r=0$, then $r^2=0$, so of course you get that $r^2=0$.

However, I don't understand why you think you get $\sin(2\theta)=0$. If $r^2=0$, then $-4\sin(\theta)=0$. That means that $\sin(\theta)=0$; where did that $2$ come from?

If you set $\theta=0$ instead, then $\sin(\theta) = \sin(0)$. How much is $\sin(0)$? How much is that when multiplied by $-4$? And what is the (only) value of $r$ that will make $r^2 = -4\sin(0)$ true?

Again, I don't understand why you think you will get "Does not exist" if you plug in $\theta=0$. This is simply not the case. (Though, if you had $r^2 = -4\cos(\theta)$, and tried to find a real value of $r$ for the case $\theta=0$, then you would be unable to find one; are you sure you are computing $\sin(0)$ correctly?)

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1) It is right.

2) You do not get DNE, you get 0... (Sqrt[-4*0] = 0)

It comes out as a very nice figure 8 loop. You seem to have understood the question, but are lacking some basic algebra/trig that the problem was not designed to test.

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I think the problem was I didn't pay attention when I was doing this problem and confused myself. Thanks for the help. – Ryan May 22 '11 at 0:19

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